Chapter 12.8, Problem 20WE

To determine

Find the perimeter.

Answer to Problem 20WE

About 25 units.

Explanation of Solution

Given:

One angle of a triangle has measure $65°$ . The side opposite this angle has length 10.The area of the triangle is 20.

Formula Used:

Law of Cosine:

  $b^2=a^2+c^2-2ac\cos B$

Law of Sine:

  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

  

Calculation:

Represent the situation and solve :

  

The area of the triangle is 20

  $\begin{array}{l}\frac{1}{2}ac\sin 65°=20\\ \Rightarrow ac=\frac{40}{\sin 65°}=44.14\\ \Rightarrow 2ac=88.28\end{array}$

Use Law of Cosine:

  $\begin{array}{l}{b}^2=a^2+c^2-2ac\cos B\\ \Rightarrow {10}^2=a^2+c^2-2ac\cos 65°\\ \Rightarrow {10}^2=a^2+c^2-(88.28)(0.4226)\\ \Rightarrow a^2+c^2=100+\mathrm{37.307....................}[\text{add 37}\text{.307 each side]}\\ \Rightarrow a^2+c^2=137.307\end{array}$

So,

  $\begin{array}{l}$ {(a+c)}^2=a^2+c^2+2ac$$ \Rightarrow {(a+c)}^2=137.307+88.28$$ \Rightarrow {(a+c)}^2=225.587$\Rightarrow a+c=\sqrt{225.587}\mathrm{................}[\text{take square root each side]}\\ \Rightarrow a+c=15.02\approx 15\\ \Rightarrow a+b+c\approx 15+10=25\end{array}$

So, the perimeter is about 25 units.